Al(s) + HCl(aq) AlCl3(aq) + H2(g)
Zn(s) + HCl(aq) ZnCl2(aq) + H2(g)
Consider the unbalanced equations above. A 0.250 g sample of a mixture of aluminum and zinc metal is reacted with an excess of hydrochloric acid. Both metals react with hydrochloric acid. The sample produces 210.2 mL of hydrogen gas at STP.
(a) What is the mass percent aluminum in the sample?
(b) What is the mass percent zinc in the sample?
To solve this problem, we need to calculate the mass percent of aluminum and zinc in the sample.
First, we need to determine the moles of hydrogen gas produced in the reaction. We can use the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.
Since the volume is given at STP (Standard Temperature and Pressure), we can assume the pressure is 1 atm, and the temperature is 273 K.
Using the given volume of hydrogen gas (210.2 mL) and converting it to liters (1 L = 1000 mL), we have:
V = 210.2 mL ÷ 1000 mL/L = 0.2102 L
Now, with the equation PV = nRT, we can solve for the number of moles of hydrogen gas (n):
n = (P * V) / (R * T)
= (1 atm * 0.2102 L) / (0.0821 L*atm/(mol*K) * 273 K)
= 0.0090 moles
Since the balanced equations show that 1 mole of aluminum reacts to produce 3 moles of hydrogen gas, and 1 mole of zinc reacts to produce 1 mole of hydrogen gas, we can determine the moles of each metal in the sample.
For aluminum:
n(Al) = (0.0090 moles H2) / (3 moles H2/1 mole Al) = 0.0030 moles Al
For zinc:
n(Zn) = (0.0090 moles H2) / (1 moles H2/1 mole Zn) = 0.0090 moles Zn
Now, we can calculate the mass of each metal in the sample.
For aluminum:
m(Al) = n(Al) * molar mass(Al)
m(Al) = 0.0030 moles * 26.98 g/mol = 0.08094 g
For zinc:
m(Zn) = n(Zn) * molar mass(Zn)
m(Zn) = 0.0090 moles * 65.38 g/mol = 0.58842 g
Finally, we can calculate the mass percent of aluminum and zinc in the sample.
(a) Mass percent of aluminum:
mass percent(Al) = (mass(Al) / mass of sample) * 100
= (0.08094 g / 0.250 g) * 100
= 32.4%
(b) Mass percent of zinc:
mass percent(Zn) = (mass(Zn) / mass of sample) * 100
= (0.58842 g / 0.250 g) * 100
= 235.4%
Therefore, the mass percent aluminum in the sample is 32.4%, and the mass percent zinc in the sample is 235.4%.